A) 2.34 g
B) 1.17 g
C) 9.36 g
D) 4.68 g
Correct Answer: D
Solution :
Key Idea (i)\[\frac{N}{{{N}_{0}}}{{\left( \frac{1}{2} \right)}^{n}}\]where n = number of half-life period \[{{N}_{0}}=\] initial concentration N = concentration left after n half-life periods. (ii) \[T=n\times {{t}_{1/2}}\] where, T = total time \[{{t}_{1/2}}=\] half-life period Here, \[{{t}_{1/2}}=3h,\,\,\,T=18h\] \[\therefore \] \[n=\frac{18}{3}=6\] Now using, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{N}_{0}}=300g\] \[\frac{N}{300}={{\left( \frac{1}{2} \right)}^{6}}\] \[\frac{N}{300}=\frac{1}{64}\] \[N=\frac{300}{64}\] N = 4.68 g
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